An isosceles triangle is a triangle that has two equal sides.
The isosceles triangle theorem states the following:
Isosceles Triangle Theorem
In an isosceles triangle, the angles opposite to the equal sides are equal. Conversely, if the base angles of a triangle are equal, then the triangle is isosceles. \(_\square\)
Consider isosceles triangle \(\triangle ABC\) with \(AB=AC,\) and suppose the internal bisector of \(\angle BAC\) intersects \(BC\) at \(D.\) Now consider the triangles \(\triangle ABD\) and \(\triangle ACD\). We have \(AB=AC\), \(AD=AD\) and \(\angle BAD=\angle CAD\) by construction. Hence, \(\triangle ABD\cong\triangle ACD\) by the SAS congruence axiom. So \(\angle ABC=\angle ACB\). \(_\square\)
Note: The converse holds, too. If we were given that \(\angle ABC=\angle ACB\), in a similar way we would get \(\triangle ABD\cong\triangle ACD\) by the AAS congruence theorem. Thus, \(AB=AC\) follows immediately.
This theorem gives an equivalence relation. In order to show that two lengths of a triangle are equal, it suffices to show that their opposite angles are equal. In fact, given any two segments \(AB\) and \(AC\) in the plane with \(A\) as a common endpoint, we have \(AB=AC\Longleftrightarrow \angle ABC=\angle ACB\). So in a geometry problem, if we are to show equality of two sides of a triangle, we can start chasing angles!
In \(\triangle ABC\) we have \(AB=AC\) and \(\angle ABC=47^\circ\). Find \(\angle BAC\).
By the isosceles triangle theorem, we have \(47^\circ=\angle ABC=\angle ACB\). Since the angles in a triangle sum up to \(180^\circ\), we have \[\angle BAC=180^\circ - \left(\angle ABC+\angle ACB\right)=180^\circ-2\times 47^\circ=86^\circ. \ _\square\]